Sharjah Indian School, Sharjah (Boys Wing) Notes on ‘Atoms and Nuclei’ [class 12] Rutherford’s model of atom According to this the entire positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus with electrons revolving around the nucleus just as planets revolve around the sun. Alpha particle scattering experiment (Geiger-Marsden Experiment) This experiment establishes Rutherford’s model of atom. Highly energetic α particles are allowed to fall a thin gold foil and the following observations were recorded: (a) A large number of α particles are found to be undeflected. (b) Some α particles are scattered at large angles. (c) A few particles are found to bounce back. The following graph illustrates the variation of angle of scattering with the number of α particles.
Conclusions (a) Atoms contain lots of empty space. (b) The entire positive charges of an atom are concentrated at the centre called nucleus. Distance of closest approach It is the closest distance between the α particle and the nucleus. It gives the approximate measure of the size of the nucleus. It can be calculated as follows: At the distance of closest approach, the K.E of an α particle gets completely converted into electrostatic potential energy. 1 2𝑒 (𝑧𝑒 ) i.e. ½ mv2 = 4𝜋𝜀 . Rearranging this, we get the distance of closest approach as, 𝑟 0
𝑧𝑒 2
r = 𝜋𝜖
0 𝑚𝑣
2
The value of r is found by this method approximately to be 10 -15m. Impact parameter (b) is the perpendicular distance between the direction of an α particle and the axis passing through the centre of the nucleus. It is given by b =
𝑍𝑒 2 cot (𝜃/2) 4𝜋𝜖 𝑜 𝐸
Limitation of Rutherford’s model:- This model could not explain the stability of atoms, despite of the acceleration of electrons around the nucleus. Page 1
BOHR ATOM MODEL Postulates:- (i) An electron in an atom could revolve in certain stable orbits without the emission of radiant energy. (ii) Electrons revolve around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is, L = nh/2π (iii) When an electron jumps from higher energy level (E2) to a lower energy level (E1), a photon is emitted having energy equal to the energy difference between the initial and final states. i.e., h ν = E2 – E1 Hence the frequency of the emitted photon is given by = (E2 – E1)/ h. It is called ‘Bohr frequency’.
Expression for the energy of an electron in the nth orbit of a Hydrogen atom. Consider an electron revolving in the nth orbit of an electron in the orbit of radius rn with a speed vn. The centripetal force required for the revolution of the electron is provided by the nuclear force of attraction. 2
So,
mvn 1 (e)(e) Thus we get, rn 4 0 rn 2
rn =
e2 4 0 mvn
--------- (1)
2
By Bohr’s second postulate, the angular momentum of the electron is given by, mvnrn = nh/2π Using equation (1), mvn
e2 4 0 mvn
2
= nh/2π This gives the velocity of the electron as,
vn
e2 . ---- (2) 2 0 nh
It implies that vn
1 n
Substituting (2) in equation (1) and solving we get, Radius of the orbit
rn =
0n2h2 -----(3) me 2
i.e.,
rn n2
Hence kinetic energy of the electron is given by, K.En = ½ mvn2
e2 2 ) = ½ m( 2 0 nh K.En =
me 4
8 0 n 2 h 2 1 (e)(e) Potential energy of the electron is given by, P.En = 4 0 rn 2
Substituting the value of radius from equation (3) we get P.En =
----------(4)
me 4 4 0 n 2 h 2 2
---------- (5)
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Finally, the total energy of the electron is En = K.En + P.En = En =
me 4
+ (
8 0 n 2 h 2 2
me 4 4 0 n 2 h 2 2
)
me 4 8 0 n 2 h 2 2
Putting the values of m,e,o and h and solving we get En =
13.6 eV n2
Note:- from the above equations it can be shown that (i) K.En = - En (ii) P.En = 2 (En) and (iii) K.En = - 2 (P.En)
Rydberg – Balmer formula Using the third postulate we get h ν = E2 – E1 =
me 4 8 0 n2 h 2 2
2
(
me 4 8 0 n1 h 2 2
2
)
me 4 1 1 2 2 2 2 8 0 h n1 n2 1 1 1 Hence the wave number is given by, = RH 2 This is Rydberg-Balmer formula 2 n2 n1 i.e.,
hc
=
Where RH is a constant called Rydberg constant. The above equation can be used to explain ‘Atomic spectra’ – [Refer text book pages 428-429] Explanation of Second postulate using de-Broglie’s concept - [Refer note book] Limitations of Bohr model - [Refer text book pages 431-432]
NUCLEI a.m.u :- It is defined as 1/12 of the mass of C-12 atom. 1a.m.u = 1.66 x 10 -27 kg. Conversion of a.m.u into MeV. E = mc2 = (1.66 x 10 -27 )(3 x 108)2 = 931 MeV Isotopes:- Nuclei with the same atomic number but different mass numbers. E.g. 2He3 and 2He4. Isobars:- Nuclei with the same mass number but different atomic numbers. E.g. 6C14 and 7N14. Isotones:- Nuclei with the same number of neutrons. E.g. 2He4 and 1H3. Radius of nucleus:- The radius of a nucleus is found to be directly proportional to the cube root of mass number (A). i.e., R α A 1/3 or R = Ro A1/3 where Ro = 10 -15m (1Fermi) Density of nucleus:- Density (ρ) = mass / volume i.e.
(ρ) =
A 3x1.66 x10 27 A = 1018 kg/m3. 1/ 3 3 15 3 3 4 / 3 ( R0 A ) 4 (10 ) 4 / 3R
Conclusions: (i) All nuclei have the same density. i.e., the density of nucleus is independent of its mass number. (ii) the highest magnitude of density indicates that the entire matter of an atom is concentrated in the nucleus. i.e., atoms contain lots of empty space. Page 3 dept. of physics, shrajah indian school, Sharjah boys wing …dept.of physics. sharjah indian school Sharjah boys wing…dept dept. of physics, shrajah indian sc
Mass defect and Binding energy Mass defect (m) of a nucleus is defined as the difference between the actual mass of a nucleus and the total individual masses of protons and neutrons inside the nucleus. i.e, m = [Zmp + (A- Z) mn – m] where m, the actual mass of a nucleus, A, the mass number, Z, the atomic number, mp and mn, mass of a proton and a neutron respectively. This mass defect is used to provide energy, called Binding energy, to bind nucleons within the nucleus. Thus Binding Energy (B.E) = mc2. Also, the binding energy per nucleon is given by (mc2/A). It is the binding energy per nucleon, rather than the total B.E, that determines the stability of a nucleus. Packing fraction: It is the mass defect of a nucleus pen nucleon. (m / A) Variation of B.E per nucleon with mass number of nuclei. The following graph shows the variation of B.E per nucleon with mass number of different nuclei.
B.E/nucleon (MeV)
Mass number (A) Conclusions: (i) The B.E per nucleon is not distributed uniformly over different mass numbers. (ii) The nuclei with mass numbers between 30 and 170 have practically the same value of the B.E/nucleon. Fe56, being with the highest B.E/nucleon is the most stable. (iii) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged. (iv) Heavy nuclei such as Uranium possess lower B.E/nucleon and hence they are less stable. Hence when they split to form comparatively lighter fragments, gaining higher B.E/nucleon, they gain stability. This process releases energy and is called nuclear fission. (v) Lighter nuclei such as Hydrogen, also possess lower B.E/nucleon and hence they are less stable. Hence when they combine to form comparatively heavier fragments, gaining higher B.E/nucleon, they gain stability. This process also releases energy and is called nuclear fusion.
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NUCLEAR FORCES They are the strongest forces that hold nucleons inside the nucleus of an atom in a tiny space , overcoming the electrostatic force of repulsion between protons. Characteristics They are the strongest force in nature. They exhibit saturation property. i.e., a given nucleon influences only those nucleons close to it. They are charge-independent. They are short-ranged They are non-central forces. Variation of potential energy with separation between nucleons The inter-nuclear force is the negative gradient of potential energy with the distance. F = - (dU/dx). Negative potential energy indicates the force of attraction between nucleons. Hence the variation of force with distance can be plotted as shown:
Pot.energy (U)
0
Ro Separation (R)
F
R 0
Ro
RADIOACTIVITY It is the phenomenon of spontaneous emission of radiations from a nucleus when it becomes unstable. The instability arises when there is a large difference in neutron and proton numbers in the nucleus. It is discovered by Henry Becquerel. Further, Rutherford found that there are three types of radiations from the nucleus, namely, , and , with the help of the experiment depicted in the figure given alongside. Properties of , and radiations.
1. Positively charged particles Negatively charged particles 2. Equivalent to He nucleus Equivalent to an electron 3. Max. speed 1/10th that of light. Max.speed nearly equal to c. 4. Ionizing power, the highest. Ionizing power, lower than
Neutral Not a particle, but an e.m radiation Equal to c, the speed of light. Ionizing power, the least.
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5. Penetrating power, the lowest. Penetrating power, higher Penetrating power, the highest. It can be demonstrated as: α β β γ γ γ paper Al sheet Lead block 6. Can be deflected in electric and Can be deflected in electric and Cannot be deflected in both the magnetic fields. magnetic fields. fields. 7. All the three can cause change in photographic plates. . Half-Life (T½ ): It is the time taken for half of a radioactive substance to decay. Thus, the fraction left after ‘n’ half lives will be (1/2)n. Mean life (): It is the time for a radioactive substance to become e-1 times the initial quantity. Activity (R): It is the number of disintegrations per second of a radioactive substance. R=
dN dt
Units of activity: Standard unit – Becquerel (Bq) 1Bq = 1disintegration per second Bigger units: Rutherford (Rd) and Curie (Ci) 1Rd = 106 disintegrations per second 1 Ci = 3.7 x 1010 disintegrations per second The law of radioactive decay. It states that the number of radioactive atoms disintegrating per second is directly proportional to the number of atoms present in the sample at that instant.
dN N dt
i.e.,
dN N , dt
where is a constant called decay constant or
disintegration constant. The negative sign indicates that the number of atoms decreases with time.
dN Further, dt . On integrating, we get N
N
t
dN N N 0 dt 0
Where N and No are the number of radioactive atoms at t = 0 and t = t
log e N NN
loge (N/No) = t.
0
t Finally,
So,
log e N log e N 0 t
-t
N = N0 e
The above equation indicates that the number radioactive atoms decreases exponentially with time as shown in the graph. Also the above equation can be re-written in terms of activity as shown:
R = R0 e-t Hence R = N
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Relation between half life and decay constant. We have, N = N0 e-t At t = T½ , N = N0/2 Hence, N0/2 = N0 e -T½ So, 2 = e T½ Taking log, loge2 = T½ T½ = (loge2)/ Solving T1 / 2
0.693
Relation between mean life and decay constant. -t We have, N = N0 e At t = , N = e-1N0 Hence, e-1N0 = N0 e - So, 1 = = 1/ Note: Those radioactive elements whose half-life is short compared to the age of the universe (13.7 billion years) are not found in observable quantities in nature today. They have, however, been seen in the laboratory in nuclear reactions. e.g. Tritium and plutonium. ALPHA DECAY The emission of an -particle from U-238 nucleus is shown as: Thus, during -decay, a nucleus loses 2 neutrons and 2 protons. Hence mass number decreases by 4 whereas, atomic number decreases by 2. Hence in general, The disintegration energy or Q-value of the reaction is from the difference in mass of the products and the reactants of the reaction. Thus, Q = (mX-mY-mHe) c2. Expression for K.E of - particle. - particle emission can be represented as: The energy released is given by Q = (mX-mY-mHe) c2.
By the law of conservation of momentum mHe vHe = mYvY. m v So, vY He He --------------- (1) mY 2 The total energy is given by Q = 1 / 2mY vY2 1 / 2mHe v He
mHe v He Using equation (1) we get, Q = 1 / 2mY mY
2
2 1 / 2mHe v He
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m So, Q = ½ mHe vHe2 He 1 Hence ½ mHe vHe2 = mY A4 Thus, K.E of -particle = ( ).Q A
mY mHe mY
.Q
BETA DECAY A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo beta decay. During β- - decay, a neutron decays into a proton, an electron and a chargeless, massless (nearly) particle called antineutrino ()
n p + e- + Thus, the atomic number increases by 1, whereas the mass number remains the same. e.g. During β+ - decay, a proton decays into a neutron, a positron and a chargeless, massless (nearly) particle called neutrino ()
p n + e+ +
Thus, the atomic number decreases by 1, whereas the mass number remains the same. e.g.
The particles neutrino () and antineutrino () are emitted from the nucleus along with the electron or positron during the decay process. Neutrinos interact only very weakly with matter; they can even penetrate the earth without being absorbed. It is for this reason that their detection is extremely difficult and their presence went unnoticed for long. The energy of β particles is not discrete, rather that continuous. This is because, the energy is being shared with the neutrinos. Hence the variation of energy of β particles can be shown as: GAMMA DECAY During gamma decay, there is no change in mass number or atomic number. During α or β decays, the nucleus will be left excited. It returns to the ground state by the emission of rays, with energy equals to the difference in the energy levels. E.g. By beta emission, the 60Co27 nucleus transforms into nucleus in its excited state. The excited 60Ni28 nucleus further de-excites to its ground state by successive emission of gamma rays, as illustrated in the energy level diagram given beside.
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Uses of Radioactivity The radiations from radioactive isotopes are used in (a) medical field - for diagnosis and treatment of diseases like goiter, cancer etc. (b) industry – to detect leakage in pipes, to monitor the uniformity of the thickness of paper in manufacturing of paper, purification of packed foods etc. (c) agriculture – to produce high-yielding varieties of crops by genetic mutation, to study the metabolic processes in plants by tracer techniques etc. (d) archeology – to find the age of fossils etc. by ‘carbon dating’ Note: Cloud chamber and Geiger-Muller counter are the two instruments used to detect the presence of radiations from a radioactive nuclei. NUCLEAR FISSION It is the process of splitting up of a nucleus into fragments, with the release of a large amount of energy. When a neutron bombards on a uranium target, the uranium nucleus breaks into two nearly equal fragments, releasing great amount of energy. A few examples of fission process are given below: and
The energy is released due the difference in the masses of products and reactants (E = mc2). Uncontrolled chain reaction is used in atom bomb whereas; controlled chain reaction takes place in a nuclear reactor. NUCLEAR REACTOR Terms: (a) Enrichment:- conversion of non-fissionable material to fissionable. (b) Criticality:- The rate of release of energy at every stage of a chain reaction remains the same. This can be done by ensuring the same number of neutrons causing fission at every stage. (c) Thermal neutrons:- the slow moving neutrons ideal for producing fission. [Note: Read the pages 454 and 455 of the text book carefully. Learn: (i) Parts of a nuclear reactor with example for each part, (ii) characteristics of good moderator, (iii) Hazards of nuclear wastes etc.] NUCLEAR FUSION It is the process of release of energy by the combination of lighter nuclei into comparatively heavier nucleus. Since the process requires tremendous amount of heat, nuclear fusion is also referred to as ‘thermo-nuclear reaction’. Answer the following: 1. Explain the mechanism of release of energy in stars- e.g. proton-proton cycle. 2. Why the nuclear fusion reaction cannot be carried out in controlled manner? 3. What are the advantages of nuclear fusion over nuclear fission? [Read text book pages 456 and page 457, for the above answers and for more examples and short questions based on nuclear fusion] Page 9
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